How to prove that $C(a)=C(a^3)$ when $|a|=5$ I am having trouble proving the following result:
"Suppose $a$ belongs to a group and $|a|=5$.  Prove that $C(a)=C(a^3)$."
($C(a)$ denotes the centralizer of $a$)
The typical way to do this would be to show that $C(a) \subseteq C(a^3)$ and $C(a^3) \subseteq C(a)$.  The first direction is easy: Let $g \in C(a)$ be arbitrary.  Then $ga^3=aga^2=a^2ga=a^3g$, so $g \in C(a^3)$ and $C(a) \subseteq C(a^3)$.
I'm having a hard time with the other direction, including seeing how $|a|=5$ works into it.
The problem also asks to find an element $a$ from some group such that $|a|=6$ and $C(a) \ne C(a^3)$.  This makes me curious: for an element $a$, is it true that $C(a)=C(a^i)$ when $\gcd(|a|,i)=1$, and thus $|a|=|a^i|$?
I'd appreciate a HINT on how to go about proving the original problem, and the generalization if it is indeed true.
Thanks.
 A: Both $a$ and $a^3$ generate the cyclic subgroup of order 5 to which they both belong. Try writing $a$ in terms of $a^3$; i.e., if $b=a^3$, express $a$ in terms of $b$. We can do this because the powers of $a$ in the problem statement are relatively prime to $5$ (and thus generate the cyclic subgroup $\langle a \rangle$.
A: Have you used that $$|a^3|=\frac{|a|}{\gcd(|a|,3)}= |a|\text{ ? }$$
Suppose now that $xa^3=a^3x$. Note that $(a^3)^2=a$ Then $$xa^3a^3=a^3xa^3$$
Can you finish?
SPOILER 

$$xa=xa^6=xa^3a^3=a^3xa^3=a^3a^3x=a^6=ax$$

As per your curiosity: one can prove that $$|a^k|=\frac{|a|}{(|a|,k)} $$
Note we have $|a|=|a^k| \iff \langle a\rangle=\langle a^k\rangle$, and from the above, $\iff (|a|,k)=1$, and as anon succinctly commented: "$x$ commutes with $a$  if and only if $x$ commutes with every power of $a$ if and only if $x$ centralizes $\langle a\rangle$. Hence $C(a)=C(b)$ whenever $a$ and $b$ generate the same cyclic subgroup."  
A: I will complete the proof, by proving two separate statements. The first being, $C(a)\subseteq C(a^3)$. Naturally, the second statement is $C(a^3) \subseteq C(a)$. First, suppose some $x \in C(a)$. This implies $ax=xa$. Acting on the right by $a^2$ gives $a^3x=a(ax)a=axa^2$. Associating again, $(ax)a^2=xa^3$. Hence, $a^3x=xa^3$. As a result $x\in C(a^3)$. Now, allow $x\in C(a^3)$. This implies that $a^3x=xa^3.$ Acting on the left by $a^3$ gives $a^6x=(a^3x)a^3=ax^6$ After considering the order of a one has $ax=xa$. Therefore $x\in C(a)$. $\square$ 
